Let \(X\cong PSL_{2}(q)\) . Then \(\frac{k+1}{(k+1,(2,q-1)f)}\mid \left|X_{x}\right|\) , as \(\left|\mathrm {Out(}X\mathrm {)}\right|=(2,q-1)f\) , by Lemma REF . Moreover, \(q\ne 4,5,9\) , since \(PSL_{2}(4)\cong PSL_{2}(5)\cong A_{5}\)  and \(PSL_{2}(9)\cong A_{6}\) , and both these
cases are ruled out in Lemma 3.5 of [1]}.
Assume that \(k+1\)  divides \((2,q-1)f\) . Thus, \(k^{2}<(2,q-1)^{2}f^{2}\) . Moreover, \(q>11\)  and \(f\geqslant 2\) , as \(\mathcal {D}\)  is non-trivial. Hence \(q+1<(2,q-1)^{2}f^{2}\) , since \(q+1\)  is the minimal primitive permutation
representation of \(X\)  for \(q>11\)  by [2]}, Theorem 5.2.2. So, \(q=3^{3} \)  and \(3\leqslant k\leqslant 5\) . Actually, \(k=5\)  as \(k+1\)  must divide \((2,q-1)f \) . However, this case cannot occur, since \(PSL_{2}(3^{3})\)  has no
transitive permutation representations of degree less then 28.
Assume that \(\frac{k+1}{(k+1,(2,q-1)f)}>1\) . Let \(U\)  be a Sylow \(u\) -subgroup
of \(X_{x}\) , where \(u\)  is a prime divisor of \(\frac{k+1}{(k+1,(2,q-1)f)}\) ,
and assume that \(U\)  fixes a further point \(y\)  of \(\mathcal {D}\) . So, \(\left|y^{X_{x}}\right|\)  is coprime to \(u\) . This is a
contradiction, since \(\frac{k+1}{(k+1,(2,q-1)f)}\)  divides \(\left|y^{X_{x}}\right|\)  by Lemma REF . Therefore \(x\)  is the unique
point fixed by \(U\)  and hence \(N_{X}(U)\leqslant X_{x}\) . Then \(X_{x}=N_{X}(U)\)  and
either \(X_{x}\cong U:Z_{\frac{q-1}{(2,q-1)}}\) , or \(X_{x}\cong D_{\frac{2(q+\epsilon )}{(2,q-1)}}\)  and \(\epsilon =\pm 1\) , according to whether \(u=p\) 
or \(u\ne p\) , respectively, by [3]}, Satz 8.5.(a).
Assume that \(X_{x}\cong U:Z_{\frac{q-1}{(2,q-1)}}\) . Then \(k^{2}=\left[X:X_{x}\right] =q+1\)  and hence \(k=3\)  and \(q=8\)  by [4]}, A5.1, since \(k \geqslant 3\) .
However, this case is ruled out in Lemma 3.6 of [1]}, since \(PSL_{2}(8)\cong \)  \(^{2}G_{2}(3)^{\prime }\) .
Assume that \(X_{x}\cong D_{\frac{2(q+\epsilon )}{(2,q-1)}}\) , where \(\epsilon =\pm 1\) . Then \(k^{2}=\frac{q(q-\epsilon )}{2}\) . If \(q\)  is even, then \(q/2=2^{f-1}\) , with \(f\)  odd, and \(z^{2}=2^{f}-\epsilon \)  for some positive
integer \(z\) , since \(k^{2}=\frac{q}{2}(q-\epsilon )\) . Also \(f>1\) , since \(X\) 
is simple, and hence \(f\geqslant 3\) . Then \(\epsilon =-1\)  and \(f=3\)  by [4]},
A3.1 and A5.1. Thus \(q=8\)  and \(k=6\) , and hence assertions (1) and (2) follow
from Theorem 3.10 of [1]}, since \(PSL_{2}(8)\cong \)  \(^{2}G_{2}(3)^{\prime }\) .
Assume that \(q\)  is odd. Then \(q\)  is a square and \(q^{1/2}\)  is the highest
power of \(p\)  dividing \(k\) . Moreover, we saw that \(q>9\) . Clearly, \(k+1\)  divides
\(\frac{q(q-\epsilon )}{2}-1\)  and hence \((q-2\epsilon )\frac{(q+\epsilon )}{2}\) . By Lemma REF (3), \(k+1\)  divides \(\left|G_{x}\right|\) , which is
\(2f\left( q+\epsilon \right) \) , and hence it divides \(\frac{(q+\epsilon )}{2}f_{p^{\prime }}\) , as \(p\mid k\)  and \(q\)  is odd. Then there is an integer \(c\geqslant 1\)  such that
\(2c(k+1)=\left( q+\epsilon \right) f_{p^{\prime }}\text{.} \) 
From (REF ) we derive that \(q^{1/2}\)  divides \(2c-\epsilon f_{p^{\prime }}\) , as \(q^{1/2}\mid k\) .
Note that \(2c>f_{p^{\prime }}\) . Indeed, if \(2c\leqslant f_{p^{\prime }}\)  then \(k\geqslant q+\epsilon -1\)  and hence \(\left( q+\epsilon -1\right) ^{2}\leqslant \frac{q(q-\epsilon )}{2}\) , which has no solutions for \(q > 9\) . Then \(2c-\epsilon f_{p^{\prime }}=\theta q^{1/2}\)  for some \(\theta \geqslant 1\) , and (REF )
becomes \(\left( \theta q^{1/2}+\epsilon f_{p^{\prime }}\right) k+\theta q^{1/2}=qf_{p^{\prime }}\) . Therefore, we obtain \(k=q^{1/2}\frac{q^{1/2}f_{p^{\prime }}-\theta }{\theta q^{1/2}+\epsilon f_{p^{\prime }}}\) .
If \(\epsilon =1\) , then \(k<q^{1/2}f_{p^{\prime }}\)  and hence \(k^{2}=\frac{q(q-1)}{2}\)  implies \(q-1<2f^{2}\) , which has no solutions for \(q>9\) .
Therefore, we have \(\epsilon =-1\)  and \(k=q^{1/2}\frac{q^{1/2}f_{p^{\prime }}-\theta }{\theta q^{1/2}-f_{p^{\prime }}}\) . We have seen that \(q-1\geqslant 2f^{2}\)  as \(q>9\) . Then \(q^{1/2}>f_{p^{\prime }}\)  and hence \(\theta q^{1/2}-f_{p^{\prime }} > (\theta -1)q^{1/2}\) . So, \(k < q^{1/2}\frac{q^{1/2}f_{p^{\prime }}-\theta }{\left( \theta -1\right) q^{1/2}}<q^{1/2}f_{p^{\prime }}\)  for \(\theta >1\) , and we obtain a
contradiction as above. Hence, \(\theta =1\) . Then \(k=q^{1/2}\frac{q^{1/2}f_{p^{\prime }}-1}{q^{1/2}-f_{p^{\prime }}}\)  which, substituted in \(k^{2}=\frac{q(q+1)}{2}\) , yields \(f_{p^{\prime }}=\frac{q^{2}+q+2}{\sqrt{q}\left( q+3\right) }\) . However, such value of \(f_{p^{\prime }}\)  is not an
integer as \(q\)  is odd and \(q>1\) . This completes the proof.
